0.05% of NaF has concentration of F
**Core Concept**
The question is about calculating the concentration of fluoride (F) in a solution of sodium fluoride (NaF). **Molarity** and **millimoles** are key concepts here, as we need to understand the relationship between the percentage concentration of NaF and the concentration of F ions.
**Why the Correct Answer is Right**
To find the concentration of F, we need to know that NaF is a strong electrolyte that completely dissociates into Na+ and F- ions in solution. Given that 0.05% of NaF is equivalent to 0.05 g/100 mL, and the molecular weight of NaF is approximately 42 g/mol, we can calculate the molarity of NaF and thus the concentration of F ions.
**Why Each Wrong Option is Incorrect**
**Option A:** This option is incorrect because it does not accurately reflect the molar concentration of F ions in a 0.05% NaF solution.
**Option B:** This option is also incorrect as it misrepresents the calculation of F ion concentration from the given percentage of NaF.
**Option C:** Similarly, this option is wrong due to an error in calculating the concentration of F ions from the percentage concentration of NaF.
**Option D:** This option is incorrect because it does not correctly calculate the concentration of F ions based on the dissociation of NaF in solution.
**Clinical Pearl / High-Yield Fact**
It's crucial to remember that the concentration of F ions is directly related to the concentration of NaF due to complete dissociation. This relationship is vital in dental and medical applications where fluoride concentrations are critical.
**Correct Answer:** D. 2.2 ppm